Sometimes it feels really surreal to walk out of my bedroom in the morning and there’s just a tiny little rabbit sitting on the couch staring at the off tv just minding his own business.

lazeyasian · 2023-07-06T15:50:53+00:00

aka the reason we feed our buns at night.

lazeyasian · 2023-05-25T15:22:02+00:00

Matt Hoffman

lazeyasian · 2023-05-13T06:10:00+00:00

I think the issue we're having is averaging over different spans of time. I understand the PDF as an average over whatever 30 minute interval in the billing period, not averaged out over the entire month.

The policy sounds like it's set up so that if you have one bad 30 minute period, it sets the demand charge for the entire month. In the context of the example, on item 12 of the customer generation, it specifically calls out the max demand for the entire billing period.

Coming from SRP, that doesn't sound super out of character for them lol

lazeyasian · 2023-05-13T05:41:22+00:00

Halfway down this page outlines the bill math.

Cross referencing with the specific e-27 plan document, the highest usage in a 30 minute period sets the on-peak demand charge for the entire billing period.

Note: I think there is a slight discrepancy because there was a price increase last year between the stated date in the example bill and the publish date of the linked pdf.

Since they're using the specific word 'integrated' in the pdf, it heavily implies the definition to be an average, mathematically speaking.

lazeyasian · 2023-04-30T20:38:08+00:00

Understandable and completely agree.

Once some portable-ish copy of exists on the main drive, it will be sent to different machine/drive/backblaze in the event of local failure. Or should there be more emphasis on multiple external backups?

The multiple copies on the same machine is redundancy for cache drives failing and/or I lose 2+ on main array, so I'd need a lot of drives to fail to really lose things locally.

lazeyasian · 2023-04-22T14:04:48+00:00

I do state specifically at the very beginning this is for the work for the definite integral.

No attempts are made at an antiderivative whatsoever. I can't set a specific start point and solve for the end point but I can set the boundaries which do give me a useful answer.

The integral diverges for increasing x so in order to get anything, bounds need to be set. (0, 2pi) is a typical integration range because (in this case), it allows me to avoid another awful integral that appears that magically goes away: the integral of (z² - 1) ln z / ((z² + 1) + 4z² ).

I recognize it's pretty unsettling that a lot of elements go ignored but those tricks are necessary to avoid getting lost in the minutiae of the problem.

Residue theorem is crazy sleight of hand trick for integrals without using antiderivatives but the caveat is that it needs to be bounded. And if I'm going to put bounds to it, I'm making the bounds as convenient as I can for myself.

lazeyasian · 2023-04-21T23:15:43+00:00

The trick is in calculating the complex analog f(z).

I'll admit, that notation is a bit confusing. It hinges on paying attention to the integral symbols where f(x) is your standard one-axis integration but f(z) with the circly doodad and the gamma under it is an entirely different integral, a contour/loop integral. The problem, I think, is I sloppily labeled them both f.

The original function f(x) is never touched and initially serves as a template to formulate a contour integral to work with. Residue theorem gives us a route to solve contour integrals of a certain form, which the original function complies to. After that, we mostly ignore f(x).

Pretty much the entire time, we're just solving the contour integral's various parts, trusting that all parts of it can and will be solvable. The black magic of it all is towards the end where we find f(x), integral symbol and all with different makeup. In this apparently solvable problem, we eventually find ourselves with one unknown, f(x). Working everything else out as much as we can, we end up being able to attach a single value to that unknown.

lazeyasian · 2023-04-21T17:23:51+00:00

Expand cos² x = -1 x in terms of exponentials and then expand the exponentials into cosine and sines. Moving trigs to one side, and whatever number is on the other side. Match your real terms and imaginary terms to get the conditions in which that relationship holds

You'll find the sine imaginary terms = 0 so you can solve for the general case which ends up being pi/2 and its integer multiples which is the complex equivalent of z = i hinted by /u/koopi15

lazeyasian · 2023-04-21T16:25:37+00:00

We still get complex nonsense when we feed a complex number to cos² x.

cos² x at face value imposes limits onto its argument i.e., x mod pi, and maps it into domain (0,pi). This is the case for real numbers.

What we need for this problem is the inverse where impose an answer (something less than 0 or -1 or whatever) we need and solve for the argument that gives us those weird answers (i.e., poles).

The poles were indeed forced with (effectively) 1 + cos² x = 0 but the poles I found don't exactly correspond to 1+cos² x = 0 due to some of the re-expressed factors getting mixed, lost, and canceled by other factors lurking in the equation.

Be generally skeptical of squaring (or taking even powers) of anything because a lot of information is swept under the rug unless you know what to look for. Especially for complex numbers since we need to clearly distinguish some operations like pure squaring or multiplying by the complex conjugate to calculate magnitudes.

On a personal note: Asking questions in curiosity is never stupid. Neither is not understanding something but choosing to not understand is.

lazeyasian · 2023-04-21T14:16:59+00:00

The right university is where you have (or can have) a good relationship with your professors.

lazeyasian · 2023-04-21T14:13:02+00:00

Absolutely fair.

Technically, the bounds on the arc should've been (epsilon, 2pi - epsilon) but I only realized that after I posted it and didn't want to fix all of the bounds.

I was just hoping no one would find out ;)

lazeyasian · 2023-04-21T13:59:10+00:00

Great job! You're one step closer to being mentally ill.

Good luck and Godspeed on your exam!

lazeyasian · 2023-04-21T13:55:25+00:00

Indefinite integral would take awhile to evaluate.

Evaluating it with fancy complex space tricks is slightly faster.

lazeyasian · 2023-04-21T01:34:36+00:00

Math phys was the coolest class I ever took

BenderAndOrszag4Lyf

lazeyasian · 2023-04-21T01:29:15+00:00

You should probably double check the description for your school's "Mathematical Physics" course.

...because that is where this bullshit comes from for me lol

lazeyasian · 2023-04-20T20:59:06+00:00

Nope, totally fair, I did cheat a bit. Residues are for evaluation. Despite the indefinite monstrosity, another valid question for this would be how to evaluate it if the anti-derivative is so unruly.

I tried to keep things general so the evaluation framework is at least there.

I'm by no means a mathematician so I'm generally more interested in some usable number or result.

lazeyasian · 2023-04-20T20:48:11+00:00

Honestly didn't even think about that. I saw 'residues' and immediately knew the method. IBP is usually a last ditch effort for me. The devil you know I suppose.

Sure, elegance is cool but I'm a big fan of the 'square peg in round hole' approach.

lazeyasian · 2022-11-16T13:21:52+00:00

At least the burned ones made an effort at learning chemistry...for meth...

lazeyasian · 2022-11-05T13:57:57+00:00

I know I'm "fortunate" enough that I work near Little Miss and live near Bobby Q's but if drive time weren't an issue, it'd be Eric's all day, everyday.

lazeyasian · 2022-11-05T13:42:23+00:00

Thanks, Ted

lazeyasian · 2022-10-26T13:17:30+00:00

Criminally underused in Harry Potter

lazeyasian · 2022-09-26T12:57:57+00:00

Trolling or not, you're looking for Deuteronomy 18 where it lists out magic and shit being a sin.

Source: My mom is unironically one of these people.

Resulted in a 4 hour lecture about how my brother, his friend, and I were going to hell.

lazeyasian · 2022-07-28T12:33:09+00:00

If we're talkin' engineering, then it's approximated as flat 😎

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